Cosine Double angle formula

Formula

$\large \cos{2\theta} \,=\, \cos^2{\theta}-\sin^2{\theta}$

Other forms

Cosine of double angle rule is expressed in some standard forms.

Angle Double angle Formula
$x$ $2x$ $\cos{2x}$ $\,=\,$ $\cos^2{x}-\sin^2{x}$
$A$ $2A$ $\cos{2A}$ $\,=\,$ $\cos^2{A}-\sin^2{A}$
$\alpha$ $2\alpha$ $\cos{2\alpha}$ $\,=\,$ $\cos^2{\alpha}-\sin^2{\alpha}$

The angle of right angled triangle can be represented by any symbol and the double angle becomes twice of that symbol. The property of the cosine of double angle is expanded in terms of cosine and sine of angle in the same form.

Proof

Consider theta is an angle of the right angled triangle and the double angle is written as $2\theta$. The cosine of double angle is expressed as $\cos{2\theta}$ mathematically. The expansion of the cos of double angle is derived in geometrical approach.

01

According to ΔICG

double angle triangle

The angle of the $\Delta ICG$ is $2\theta$. Express cosine of double angle in terms of the length of the sides.

$\cos{2\theta} \,=\, \dfrac{CI}{CG}$

The side $\overline{CD}$ is divided into two sides $\overline{CI}$ and $\overline{ID}$ at point $I$. Hence, the length of the side $\overline{CI}$ can be written as the subtraction of length of the side $\overline{ID}$ from length of the side $\overline{CD}$.

$\implies \cos{2\theta} \,=\, \dfrac{CD-ID}{CG}$

$\implies \cos{2\theta} \,=\, \dfrac{CD}{CG}-\dfrac{ID}{CG}$

02

According to ΔDCF

right angled triangle DCF

$\overline{CD}$ is the adjacent side of the right angled triangle $DCF$. Transform the length of the side $\overline{CD}$ by considering this triangle.

$\cos{\theta} \,=\, \dfrac{CD}{CF}$

$\implies CD = CF \cos{\theta}$

Now, replace the length of the side $\overline{CD}$ by the product of the length of the side $\overline{CF}$ and cosine of angle theta.

$\implies \cos{2\theta} \,=\, \dfrac{CF \cos{\theta}}{CG}-\dfrac{ID}{CG}$

03

According to ΔFCG

$\implies \cos{2\theta} \,=\, \dfrac{CF}{CG} \times \cos{\theta}-\dfrac{ID}{CG}$

right angled triangle FCG

$\overline{CF}$ and $\overline{CG}$ are the sides of the $\Delta FCG$. According to this triangle, the ratio of length of the side $\overline{CF}$ and the length of the side $\overline{CG}$ is cosine of the angle theta.

$\cos{\theta} \,=\, \dfrac{CF}{CG}$

The ratio of the sides can be replaced by the cosine of the angle theta.

$\implies \cos{2\theta} \,=\, \cos{\theta} \times \cos{\theta}-\dfrac{ID}{CG}$

$\implies \cos{2\theta} \,=\, \cos^2{\theta}-\dfrac{ID}{CG}$

04

Parallelism of sides ID and FH

double angle triangle

$\overline{ID}$ and $\overline{FH}$ are sides but they are parallel lines. Therefore, the length of the side $\overline{ID}$ is exactly equal to the length of the side $\overline{FH}$.

$ID \,=\, FH$

Now, replace the length of the side $\overline{ID}$ by the length of the side $\overline{FH}$.

$\implies \cos{2\theta} \,=\, \cos^2{\theta}-\dfrac{FH}{CG}$

05

According to ΔFGH

right angled triangle HGF

$\overline{FH}$ is a side of the right angled triangle $FGH$. According to this triangle, express the length of the side $\overline{FH}$ in alternative form.

$\sin{\theta} \,=\, \dfrac{FH}{FG}$

$\implies FH \,=\, FG \sin{\theta}$

Replace the length of the side $\overline{FH}$ by the product of the length of the side $\overline{FG}$ and sine of angle theta in the expansion of the cosine of the double angle.

$\implies \cos{2\theta} \,=\, \cos^2{\theta}-\dfrac{FG \sin{\theta}}{CG}$

06

According to ΔFCG

$\implies \cos{2\theta} \,=\, \cos^2{\theta}-\dfrac{FG}{CG} \times \sin{\theta}$

right angled triangle FCG

$\overline{FG}$ and $\overline{CG}$ are sides of the $\Delta FCG$. Hence, consider this triangle one more time to express the ratio of them in its equal form.

$\sin{\theta} \,=\, \dfrac{FG}{CG}$

Therefore, replace the ratio of the length of the sides $\overline{FG}$ to $\overline{CG}$ by the sine of the angle theta.

$\implies \cos{2\theta} \,=\, \cos^2{\theta}-\sin{\theta} \times \sin{\theta}$

$\therefore \,\,\,\,\,\, \cos{2\theta} \,=\, \cos^2{\theta}-\sin^2{\theta}$

Therefore, it is proved that the expansion of the cosine of double is equal to the subtraction of the square of the sine of angle from the square of the cosine of the angle.

The expansion of the cosine of double angle is used as a trigonometric formula to expand the cosine of double angle in terms of cosine and sine of angle and vice-versa.

Verification

Take angle $\theta \,=\, 30^\circ$.

Evaluate the value of cosine of double angle.

$\cos{(2 \times 30^\circ)}$

$=\,\,\, \cos{60^\circ}$

$=\,\,\, \dfrac{1}{2}$

Find the value of the expansion of the cos of double angle formula.

$\cos^2{30^\circ}-\sin^2{30^\circ}$

$=\,\,\, {\Bigg(\dfrac{\sqrt{3}}{2}\Bigg)}^2-{\Bigg(\dfrac{1}{2}\Bigg)}^2$

$=\,\,\, \dfrac{3}{4}-\dfrac{1}{4}$

$=\,\,\, \dfrac{3-1}{4}$

$=\,\,\, \dfrac{2}{4}$

$\require{cancel} =\,\,\, \dfrac{\cancel{2}}{\cancel{4}}$

$=\,\,\, \dfrac{1}{2}$

Compare both values to confirm this formula mathematically.

$\cos{(2 \times 30^\circ)}$ $\,=\,$ $\cos^2{30^\circ}-\sin^2{30^\circ}$ $\,=\,$ $\dfrac{1}{2}$

The values of both sides are same. Hence, the cos of double angle law is true for all the angles. Therefore, it is called as a double angle trigonometric identity.

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